Time varying force horizontal `F = K_(o)t` is applied on a block of mass 'm' placed at rest on a rough 8, the acceleration of the block horizontal surface `( mu` is coefficient of friction). At a time `t = (3mu mg)/(K_(0)),` is given by a = p `mu`g `mu g` where p = ...... (`K_(0)` is a positive constant)

Options

A. 2
B. 4
C. 8
D. 5

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Forces Acting on the Block
The block of mass \( m \) is subjected to a time-varying force \( F = K_0 t \) and the frictional force due to the rough surface. The frictional force can be expressed as \( f_{\text{friction}} = \mu mg \), where \( \mu \) is the coefficient of friction.

### Step 2: Apply Newton's Second Law
According to Newton's second law, the net force acting on the block is equal to the mass of the block multiplied by its acceleration:
\[ F - f_{\text{friction}} = ma \]
Substituting the expressions for \( F \) and \( f_{\text{friction}} \):
\[ K_0 t - \mu mg = ma \]

### Step 3: Substitute the Given Time
We are given that at time \( t = \frac{3 \mu mg}{K_0} \). Substitute this value into the equation:
\[ K_0 \left(\frac{3 \mu mg}{K_0}\right) - \mu mg = ma \]
This simplifies to:
\[ 3 \mu mg - \mu mg = ma \]
\[ 2 \mu mg = ma \]

### Step 4: Solve for Acceleration \( a \)
Now, we can isolate \( a \):
\[ a = \frac{2 \mu mg}{m} = 2 \mu g \]

### Step 5: Compare with the Given Acceleration
We are given that the acceleration can also be expressed as \( a = p \mu g \). Therefore, we can set up the equation:
\[ 2 \mu g = p \mu g \]

### Step 6: Solve for \( p \)
To find \( p \), we can divide both sides by \( \mu g \) (assuming \( \mu g \neq 0 \)):
\[ p = 2 \]

### Final Answer
Thus, the value of \( p \) is:
\[ \boxed{2} \]